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Math Help - Proofs of limits of sequences

  1. #1
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    Proofs of limits of sequences

    Hello,
    I have some questions about my homework. I need to prove the following limits of sequences using the limit definition:

    1. \lim_{n\to\infty} \frac{2n^2-6n+1}{3n^2+n-3} = \frac{2}{3}
    2. \lim_{n\to\infty} (\frac{2}{3})^n = 0

    my proofs look like this:
    1. for every \varepsilon > 0 we will choose N(\varepsilon) = 2 + \frac{3}{\varepsilon}, and then:
    \forall n > N(\varepsilon) \left | \frac{2n^2-6n+1}{3n^2+n-3} - \frac{2}{3} \right | = \left | \frac{-20n+9}{9n^2+3n-9} \right | \underset{\forall n\geq 1}{=} \frac{20n-9}{9n^2+3n-9} < \frac{20n+4n}{9n^2+(3n-9)} \underset{\forall n\geq 2}{<} \frac{24n}{9n^2-n^2} = \frac{3}{n} < \frac{3}{N} = \frac{3}{2 + \frac{3}{\varepsilon}} < \frac{3}{\frac{3}{\varepsilon}}= \varepsilon

    2. for every \varepsilon > 0 we will choose N(\varepsilon) = \log_{\frac{2}{3}}\varepsilon, and then:
    \forall n > N(\varepsilon) \left | (\frac{2}{3})^n - 0 \right | = (\frac{2}{3})^n \underset{\forall n > N}{<} (\frac{2}{3})^N = (\frac{2}{3})^{log_{\frac{2}{3}}\varepsilon} = \varepsilon

    I'm new to this kind of proofs and I'm not sure if this correct,
    anyone can tell me if it's the correct way to do this?
    Thank you.
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  2. #2
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    Re: Proofs of limits of sequences

    Quote Originally Posted by Kaze123 View Post
    2. \lim_{n\to\infty} (\frac{2}{3})^n = 0

    Suppose that 0<|r|<1 we prove that (|r|^n)\to 0.

    (\exists h>0)\left[\frac{1}{|r|}=1+h\right] so \left[\frac{1}{|r|^n}=(1+h)^n\ge 1+nh>nh\right]

    Thus |r|^n<\frac{1}{nh} from which we see (|r|^n)\to 0..

    Now use on your #2.
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